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Fascinating Dirichlet Beta Function Relationships

One important L-function - closely related to the Riemann zeta function - is known as Dirichlet’s beta function (also Catalan’s beta function) with its L-series (i.e. L1) sum over the integers and product over the primes expressions as follows 1/1 s – 1/3 s + 1/5 s – 1/7 s + 1/9 s –   …    = 1/(1 + 1/3 s ) *   1/(1 – 1/5 s ) * 1/(1 + 1/7 s ) * … Then in the simplest case when s = 1, we have, 1 – 1/3 + 1/5 – 1/7 + 1/9 –   …   =   3/4 * 5/4 * 7/8 * …         = π/4. So the well-known Leibniz formula for π represents a special case of the Dirichlet beta function i.e. β(s), for s = 1. And when s = 1, 3, 5, …   β(s), results in a value of the form kπ s , where k is a rational number. So for example β(3) = 1 – 1/3 3 + 1/5 3 – 1/7 3 + 1/9 3 –   …   = π 3 /32 and β(5) = 1 – 1/3 5 + 1/5 5 – 1/7 5 + 1/9 5 –   …   = 5π 5 /1536. Now this function can be connected with the Riemann zeta function in a surprising manner. For using a similar type ge
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More on Riemann Generalisation (1)

In an earlier entry “ Intertwining L1and L2 Functions ” I showed that (1/1 2 + 1/2 2   + 1/3 2 + 1/4 2 + …) 2 /(1/1 4 + 1/2 4   + 1/3 4 + 1/4 4 + …)     =  (π 2 /6) 2 /(π 4 /90)   = 90/36 = 5/2. And   that this can be written as a Dirichlet L-series i.e. 1/1 2 + 2/2 2 + 2/3 2 + 2/4 2 + 2/5 2 + 4/6 2 + 2/7 2 + 2/8 2 + 2/9 2 + 4/10 2 + … = 5/2 This in turn makes use of our generalised Riemann zeta function formula, 1/1 s + (1 + k 1 )/2 s + (1 + k 2 )/3 s + (1 + k 1 )/4 s + (1 + k 3 )/5 s + {(1 + k 1 )(1 + k 2 )}/6 s + … = 1/{1 – (1 + k 1 )/(2 s + k 1 )} * 1/{1 – (1 + k 2 )/(3 s + k 2 )} * 1/{1 – (1 + k 3 )/(5 s + k 3 )} * …,   where k 1 , k 2 , k 3 , …    = 1. So ζ(2) 2 / ζ(4) = 1/1 2 + 2/2 2 + 2/3 2 + 2/4 2 + 2/5 2 + 4/6 2 + 2/7 2 + 2/8 2 + … = 5/2 And the result can be generalised so that when s is a positive even integer, (1/1 s + 1/2 s   + 1/3 s + 1/4 s + …) 2 /(1/1 2s + 1/2 2s   + 1/3 2s + 1/4 2s + …)   =
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Dynamic Appreciation of Riemann Hypothesis

Imagine we are presented with the first of the Riemann zeros i.e. 1/2 + 14.134725 …, (or equally 1/2 –   14.134725 …), without knowing the origin of the value and asked to find a solution to the following infinite equation, k 1 – (1 /2 + 14.134725 …) + k 2 – (1 /2 + 14.134725 …) + k 3 – (1 /2 + 14.134725 …) + k 4 – (1 /2 + 14.134725 …) + …    = 0, where k 1, k 2, k 3, k 4 , … are complex numbers, then by lucky chance we might be able to find that when k 1, k 2, k 3, k 4 , … represent the natural numbers 1, 2, 3, 4 , … respectively, that the value of the equation = 0. And imagine in turn if we were presented in turn with all of the known Riemann zeros and asked to solve the same equation for each one individually, we might then find that in every case when k 1, k 2, k 3, k 4 , …, represent the natural numbers 1, 2, 3, 4 , … respectively, that again the value of the equation = 0. So this would indeed be remarkable! For on the one hand we would have a list of known
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More Connections

It struck me clearly the other day that each L2 function can be expressed as the quotient of two L1 functions. For example in the Riemann zeta function (where s = 2) the first term in the product over primes expression is 4/3, which is related to the corresponding first prime 2, through the relationship 1 – 1/ 2 2 . Now this can be expressed as the quotient of the functions, 1/1 2 + 1/2 2 + 1/3 2   + 1/5 2 + … and 1/1 2   + 1/3 2 + 1/5 2   + 1/7 2 + … (where in the latter case all numbers which contain 2 as a shared factor are omitted). So (1/1 2 + 1/2 2 + 1/3 2   + 1/5 2 + …)/(1/1 2   + 1/3 2 + 1/5 2   + 1/7 2 + …) = 1 + ( 1/2 2 ) 1 + ( 1/2 2 ) 2 + ( 1/2 2 ) 3 + …     = 4/3 And notice the complementarity here! Whereas the natural numbers represent the base aspect in the L1, they represent the dimensional aspect of number in the L2; and whereas 2 is constant as a dimensional number in the L1 it is constant (as reciprocal) in L2. So this properly signifies
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Intertwining L1 and L2 Functions

As is well known when s = 4, ζ(4) = 1/1 4 + 1/2 4   + 1/3 4 + 1/4 4 + …    = π 4 /90 and when s = 2, ζ(2) = 1/1 2 + 1/2 2   + 1/3 2 + 1/4 2 + …     = π 2 /6, so that (1/1 2 + 1/2 2   + 1/3 2 + 1/4 2 + …) 2   = π 4 /36    Therefore (1/1 2 + 1/2 2   + 1/3 2 + 1/4 2 + …) 2 /(1/1 4 + 1/2 4   + 1/3 4 + 1/4 4 + …)   = 90/36 = 5/2 However we have already seen that this can be written as a Dirichlet L-series i.e. 1/1 2 + 2/2 2 + 2/3 2 + 2/4 2 + 2/5 2 + 4/6 2 + 2/7 2 + 2/8 2 + 2/9 2 + 4/10 2 + … = 5/2 So ζ(2) 2 / ζ(4) = 1/1 2 + 2/2 2 + 2/3 2 + 2/4 2 + 2/5 2 + 4/6 2 + 2/7 2 + 2/8 2 + 2/9 2 + 4/10 2 + … Thus we have expressed the square of one Dirichlet series (the Riemann zeta function where s = 2) divided by another Dirichlet series (the Riemann zeta function where s = 4) by yet another Dirichlet series (where s = 2). And this relationship can equally be expressed as a product over primes expression. So ζ(2) 2 / ζ(4)   = (4/3 *
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Another Interesting Generalisation (2)

In the last entry, I mentioned the new general formula, which can be used to extend relationships as between sum over integers and product over primes expressions with respect to the Riemann zeta function i.e. 1/1 s + (1 + k 1 )/2 s + (1 + k 2 )/3 s + (1 + k 1 )/4 s + (1 + k 3 )/5 s + {(1 + k 1 )(1 + k 2 )}/6 s + … = 1/{1 – (1 + k 1 )/(2 s + k 1 )} * 1/{1 – (1 + k 2 )/(3 s + k 2 )} * 1/{1 – (1 + k 3 )/(5 s + k 3 )} * …,   where k 1 , k 2 , k 3 , … are rational numbers which can be either positive or negative. In that entry, I demonstrated the - surely - interesting fact that when s is 2, 4, 6, … and k 1 = k 2 = k 3, …   = 1, that the value of the two expressions (sum over the integers and product over the primes respectively) is a rational number. In this case there is a clear relationship as between the values of the standard zeta function for s and 2s respectively. So - again when s is a positive even integer - the value of the zeta expression (for both sum
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Another Interesting Generalisation (1)

It just struck recently that the sum over the integers and product over the primes expressions, with respect to the Riemann zeta function, represent a special case of a more comprehensive relationship, which can be expressed in general terms as follows: 1/1 s + (1 + k 1 )/2 s + (1 + k 2 )/3 s + (1 + k 1 )/4 s + (1 + k 3 )/5 s + {(1 + k 1 )(1 + k 2 )}/6 s + … = 1/{1 – (1 + k 1 )/(2 s + k 1 )} * 1/{1 – (1 + k 2 )/(3 s + k 2 )} * 1/{1 – (1 + k 3 )/(5 s + k 3 )} * …   where k 1 , k 2 , k 3 , … are rational numbers which can be either positive or negative. So the standard equation represents the situation where the values for  k 1 , k 2 , k 3 , …   = 0 .   It can be seen here it is the distinct prime factors of a number that determines the nature of the sum over the integers expression. So for example, 6 is the 1st composite natural number with two distinct prime factors. Therefore as we can see from the new formula we must here thereby multiply 1 + k 1
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